Question

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

Answer 1

Answer:

A. $C_5H_7N$

B. $C_4H_5N_2O$

Explanation:

Hello,

a. In this case, based on the given percentages, one computes the moles of C, H and N as follows:

$n_C=0.7403gC*\frac{1molC}{12gC}=0.062molC\\n_H=0.087gH*\frac{1molH}{1gH}=0.087molH\\n_N=0.1727gN*\frac{1molN}{14gN}=0.0123molN$

Now, we divide each element's moles by the smallest amount of moles, that is nitrogen's moles:

$C=\frac{0.062}{0.0123}=5;H=\frac{0.087}{0.0123}=7;N=\frac{0.0123}{0.0123}=1$

So the empirical formula is:

$C_5H_7N$

b. In this case, based on the given percentages, one computes the moles of C, H, N and O as follows:

$n_C=0.4948gC*\frac{1molC}{12gC}=0.0396molC\\n_H=0.0519gH*\frac{1molH}{1gH}=0.0519molH\\n_N=0.2885gN*\frac{1molN}{14gN}=0.021molN\\n_O=0.1648gO*\frac{1molO}{16gO}=0.0103molO$

Now, we divide each element's moles by the smallest amount of moles, that is oxugen's moles:

$C=\frac{0.0396}{0.0103}=4;H=\frac{0.0519}{0.0103}=5;N=\frac{0.021}{0.0103}=2;O=\frac{0.0103}{0.0103}=1$

So the empirical formula is:

$C_4H_5N_2O$

Best regards.

Answer 2

A.   
elements                              C                H          N
percentage composition    74.03          8.70       17.27
Molecular mass                   12                1          14

# of mole                            6.17              8.70        1.23

÷smallest mole                    5.0                 7.0           1.0

mole ratio                              5    :               7      ;       1

THE EMPERICAL FORMUKLA FOR A. IS  C5H7O

NOTE: #of mole = percentage composition ÷ Mr
            and the ÷ smallest mole is used to find the ratio...for the above question a. it is 1.23


and b. should be done using the same procedure